Solve left(2x-3right)left(2x-3right)-left(x+5right)left(x+5right)=-23

$$\left( 2x-3 \right) \left( 2x-3 \right) - \left( x+5 \right) \left( x+5 \right) = -23$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=7$$

Steps Using the Quadratic Formula

Multiply 2x-3 and 2x-3 to get \left(2x-3\right)^{2}.

$$\left(2x-3\right)^{2}-\left(x+5\right)\left(x+5\right)=-23$$

Multiply x+5 and x+5 to get \left(x+5\right)^{2}.

$$\left(2x-3\right)^{2}-\left(x+5\right)^{2}=-23$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9-\left(x+5\right)^{2}=-23$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$4x^{2}-12x+9-\left(x^{2}+10x+25\right)=-23$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$4x^{2}-12x+9-x^{2}-10x-25=-23$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-12x+9-10x-25=-23$$

Combine -12x and -10x to get -22x.

$$3x^{2}-22x+9-25=-23$$

Subtract 25 from 9 to get -16.

$$3x^{2}-22x-16=-23$$

Add 23 to both sides.

$$3x^{2}-22x-16+23=0$$

Add -16 and 23 to get 7.

$$3x^{2}-22x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -22 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 3\times 7}}{2\times 3}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 3\times 7}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-22\right)±\sqrt{484-12\times 7}}{2\times 3}$$

Multiply -12 times 7.

$$x=\frac{-\left(-22\right)±\sqrt{484-84}}{2\times 3}$$

Add 484 to -84.

$$x=\frac{-\left(-22\right)±\sqrt{400}}{2\times 3}$$

Take the square root of 400.

$$x=\frac{-\left(-22\right)±20}{2\times 3}$$

The opposite of -22 is 22.

$$x=\frac{22±20}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{22±20}{6}$$

Now solve the equation x=\frac{22±20}{6} when ± is plus. Add 22 to 20.

$$x=\frac{42}{6}$$

Divide 42 by 6.

$$x=7$$

Now solve the equation x=\frac{22±20}{6} when ± is minus. Subtract 20 from 22.

$$x=\frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{3}$$

The equation is now solved.

$$x=7 x=\frac{1}{3}$$

Steps for Completing the Square

Multiply 2x-3 and 2x-3 to get \left(2x-3\right)^{2}.

$$\left(2x-3\right)^{2}-\left(x+5\right)\left(x+5\right)=-23$$

Multiply x+5 and x+5 to get \left(x+5\right)^{2}.

$$\left(2x-3\right)^{2}-\left(x+5\right)^{2}=-23$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9-\left(x+5\right)^{2}=-23$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$4x^{2}-12x+9-\left(x^{2}+10x+25\right)=-23$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$4x^{2}-12x+9-x^{2}-10x-25=-23$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-12x+9-10x-25=-23$$

Combine -12x and -10x to get -22x.

$$3x^{2}-22x+9-25=-23$$

Subtract 25 from 9 to get -16.

$$3x^{2}-22x-16=-23$$

Add 16 to both sides.

$$3x^{2}-22x=-23+16$$

Add -23 and 16 to get -7.

$$3x^{2}-22x=-7$$

Divide both sides by 3.

$$\frac{3x^{2}-22x}{3}=-\frac{7}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{22}{3}x=-\frac{7}{3}$$

Divide -\frac{22}{3}, the coefficient of the x term, by 2 to get -\frac{11}{3}. Then add the square of -\frac{11}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{22}{3}x+\left(-\frac{11}{3}\right)^{2}=-\frac{7}{3}+\left(-\frac{11}{3}\right)^{2}$$

Square -\frac{11}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{22}{3}x+\frac{121}{9}=-\frac{7}{3}+\frac{121}{9}$$

Add -\frac{7}{3} to \frac{121}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{22}{3}x+\frac{121}{9}=\frac{100}{9}$$

Factor x^{2}-\frac{22}{3}x+\frac{121}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{3}\right)^{2}=\frac{100}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Simplify.

$$x-\frac{11}{3}=\frac{10}{3} x-\frac{11}{3}=-\frac{10}{3}$$

Add \frac{11}{3} to both sides of the equation.

$$x=7 x=\frac{1}{3}$$